SN1 Mechanism

The SN1 mechanism is similar to SN2 in that you get a substitution product, but the path to get there is completely different. It’s important that we understand how it’s different from SN2

Concept: Drawing the SN1 Mechanism

11m
Video Transcript

All right guys, so now we're pretty much pros at the SN2 mechanism. But it turns out that not all substitution reactions proceed through the SN2. There's actually another mechanism out there that also produces substitution at the end and that mechanism is called SN1.
If I were to just generally paraphrase what SN1 is I would say the following sentence. What you're going to notice is that quite a few of the keywords that we use for SN2 are going to change. What it is is that a neutral nucleophile, so that's a big deal, neutral, is going to react with an inaccessible leaving group.
What am I talking about here? Well, neutral means that – remember that SN2 is negative. So this one means – neutral, it's not going to have that strong nucleophilic property. Inaccessible has to do with R groups. So we're going to get there in a second, but you can already start to think, that must mean that there's a lot of R groups maybe. That's going to produce substitution in two steps.
So what you can see is that in this sentence almost every word has changed except for substitution. What that means is that at the end I still get a very similar product, which is a substitution product, but all of the conditions have pretty much changed. Now instead of being negative, it's neutral. Instead of being accessible, it's inaccessible. And instead of happening in one step, it's going to happen in two.
Let's just go ahead and start breaking this mechanism down. The biggest thing you're going to notice that's different about the SN2 versus the SN1 is that my nucleophile looks totally different. Before, my nucleophile was the initiator in the reaction. My nucleophile was like the arrow. It had a very strong negative charge and it was looking for any back side to react with.
If I draw that same first step here, that's a huge mistake. Do you know why? Because this is not negatively charged, so it's not very nucleophilic. It's not looking for a back side right now. In fact, it's happy. It's neutral. So the nucleophile is just chilling. To just draw it going to the backside would be a huge mistake. That nucleophile is chilling. It doesn't need to do anything. It's neutral.
That means that the first step must be something else. It turns out that the first step is going to be very unusual. And I know that it's going to be weird, so I'm just going to try to get it over with. It's that the X or the halogen, the leaving group, is going to take off all on its own.
This is weird because every since I've been teaching you about mechanisms, I've always said that you make a bond first and then after you break a bond. But here, what I'm doing is I'm breaking a bond without making one first. I'm just breaking it by itself. Why is that possible? How does this even make sense?
Unfortunately, I don't have another reaction in organic chemistry that I can compare this to. We've never done this before. But if you think way back, when you took gen chem two, we actually do know a reaction that was similar to this. I'll try to make the explanation quick because I know you barely remember gen chem two probably.
Remember when we were thinking back on acids and bases and we learned about titrations. And we talked about how pH had to do with the amount of hydrogen ions you had and the OH ions. What we learned is that the Kw of water, the dissociation constant of water, was 1 x 10- –do you guys remember? It's been such a long time. Negative 14. What the hell does that mean?
What that means is that water is normally H2O and that's the way it's happy. But for some random reason, one out of every 100 billion molecules is going to decide to just split apart on its own and make H+ and OH-. For a split second it's going to do that. And guess what's going to happen? It's going to hate its life because now it's two ions and it's way less stable. Then it's going to come back together.
This is a random process that's happening all the time. You could almost think of it like a divorce, but then they get right back together because they realize that they actually like each other.
In the same way, alkyl halides have a dissociation constant as well. So I'm going to put here KRX. But that dissociation constant is actually going to be a lot higher. Why? Because water is very unstable after it dissociates. But alkyl halides actually already have a really strong dipole. Then on top of that, the bond, the carbon-halogen bond is actually a very, very weak bond in comparison to the water.
What that means is that it's actually going to be easier for the alkyl halide to dissociate than water will be. I don't know the exact number. I'm just to make one up. Let's say that instead it's 10-7. Remember that this is on an exponential scale so that doesn't mean it's twice as good. That means it's like a million times as good.
All I'm trying to say is that this is going to wind up giving me ions as well, but at a better rate than I would usually get for – or a better equilibrium than I would usually get for water. What that means is I'm going to wind up getting C+ and I'm going to wind up getting X-. That's what's happening here. All I'm trying to say is that random processes are going to drive my alkyl halide to ionize and that's always going to be the first step.
What that means is that my nucleophile's chilling. It's neutral. But now, for some reason, my alkyl halides going to ionize by itself. What that's going to give me is a positive charge and a negative charge. The positive charge is what we're going to call our carbocation.
Now what our carbocation is going to look like is it's going to be this carbon right there. I'm just going to draw that carbon, but now it's only attached to three things. It's attached to an H. It's attached to an ethyl group on one side and it's attached to a methyl group on the other.
Now this carbon wants to have four bonds, but it only has three, so it needs a positive charge. Now do you guys remember what the hybridization in geometry was of a carbocation, three bonds? It should be – I'm just going to write here. It's going to be Sp2 because it only has three groups or three bond sites and it should be trigonal planar. I'm going to put TP but that's just going to mean planar, trigonal planar.
Notice that before I had my methyl on a wedge and my hydrogen on a dash. You can't draw it that way anymore because trigonal planar means it's all on the same plane. So it's actually wrong to draw it that way. Now we have to draw everything with a stick on the plane. Is that cool so far?
This is what's called our carbocation intermediate. And this brings us to the concept of transition state versus intermediate. Remember that the middle step of an SN2 was to make a transition state. Well, in this case, for an SN1, your middle step is an intermediate. What does that mean? You actually can isolate this. This is the molecule that it's not very stable, but it actually can be isolated and I can make some of it. I can make carbocations and I could keep them in a test tube and I could keep them for a little bit.
Now I have my carbocation intermediate. What the next step going to be? Is it just going to end there? Obviously not. The next step is going to be that now I have my weak nucleophile that was just chilling, but now I have a very strong electrophile. A carbocation is one of the strongest electrophiles possible. When my nucleophile sees that, you know what, it might not be negative, but it has electrons, so then it's going to go ahead and attack the carbocation. So in the second step, my nucleophile comes and attacks.
But now I have a question for you guys. Do I have a front side and a back side that I have to worry about? No, because it turns out I don't have a halogen attached anymore. Remember before I always had my halogen attached so I could only go from the back, but now all I have is this planar structure that I could either attack from the front or from the back equally. It doesn't matter which one. So that means that I have two different ways that I could attack. I could either attack from the back or I could attack from the front. And that could lead, if I have a chiral center, that could lead to two different products.
What that means is that – by the way, is this chiral? Yeah, it is. It's the same compound as before. So what that means is that I could get the example where my – basically, where my nucleophile is here, but everything's exactly the same. What that means is that the chirality doesn't change at all. That's if it let's say attacked from the front because it's just – basically, it's attacking from the same side that the X attacks from or that the X left from, so basically the X was here, my nucleophile attacks from the front, so it attaches to the same place.
But the other product that it could give me is the back product. So what it could also do is it could attack from the back side, which means that the nucleophile would be attached to where the ethyl group used to be. And what that means is that I would get the other chiral center.
Is there a way to determine exactly which one I'm going to get? No. It turns out that because this is trigonal planar, it's actually a 50/50 shot. I have 50% chance that I'm going to collide into the front. Remember this is just based on collisions. I also have a 50% chance that I'm going to collide into the back. So what that means is that I know that I'm going to have more than one enantiomer possible if it is a chiral center.
We're going to hold that thought and I'm going to keep on explaining more things about this reaction, but I just want you guys to see how that is right now.
Is there anything else that I should draw? Yes, remember that there's going to be the X and the X is going to have a negative sign.
Notice that my product came out similar to the SN2. I'm going to get substitution at the end. The only difference is that I have some differences with the arrows first of all. The arrows are completely different. And my stereochemistry seems a little bit off. So we're going to go ahead and talk about that in a little bit. 

Summary: A neutral nucleophile reacts with an inaccessible leaving group to produce substitution in two-step.

Concept: Why highly substituted leaving groups favor SN1

2m
Video Transcript

Now what I want to do is talk about the inaccessible leaving group part. That has to do with R groups and it has to do with carbocations. So basically, there's this trend about carbocation stability that says that the more R groups that you have directly attached to a carbocation, the more stable it's going to be.
It turns out that since you need carbocations in order for the nucleophile to attack it, that means that the more R groups I have on my alkyl halide, the better it's going to be. So what that means is that a tertiary alkyl halide or a tertiary halide is going to be way better at this than a primary. I'm just trying to point that out because a tertiary will be the one that's the most stable. Then this would be the side that's the most instable. It should be unstable, I'm sorry. Not instable. Oh my gosh.
Don't worry too much about this other stuff in the middle. I'm going to go through a much more robust explanation of carbocations in a little bit. But for right now just know that the more R groups, the better. The less R groups, the worse.  

The slow step of this mechanism is the formation of a carbocation intermediate. These types of intermediates are unstable, so anything that we can do to stabilize them will help them form faster.

-R groups stabilize carbocations through a phenomenon called hyperconjugation. Meaning that the more substituted the carbocation, the more stable it is. 

Concept: Understanding the properties of SN1

9m

You are a manager at Pepe and Son Carbocations Inc, and your boss has asked you to increase production of your product (carbocations in a box- these are awesome).

You have plenty of boxes, but it takes time for the conveyer belt to crank out these custom carbocations.

  • Will increasing the number of boxes increase the amount of product?
  • Will increasing the speed of the conveyer belt increase the amount of product? 

Properties of SN1 reactions:

  • Nucleophile =  Weak
  • Leaving Group =  Highly Substituted
  • Reaction coordinate = Intermediate
  • Reaction = Two-Step
  • Rate =  Unimolecular
  • Rate =  k[RX]
  • Stereochemistry = Racemic
  • Nickname = Solvolysis

Problem: Predict the product of the following reaction:

 

11m

The product must be a racemate if the original leaving group is located on a chiral center.

NOTE: Substitution reactions with neutral nucleophiles require an additional deprotonation step. 

Problem: Predict the product of the following reaction:

 

5m

SN1 Mechanism Additional Practice Problems

Complete the following mechanism for an SN1 reaction that occurs when 2-chloro-2-methylbutane is heated in water. NOTE, ALTHOUGH AN E1 PROCESS WILL ACCOMPANY THIS SN1 REACTION, WE ARE ONLY INTERESTED IN THE SN1 MECHANISM FOR THIS PROBLEM. Use appropriate arrows to show all movement of electrons, show all non-bonding electrons as dots and show any formal charges. If any of the species are really racemic mixtures of enantiomers, you only need to draw one stereoisomer and write "racemic". In the boxes provided, write which kind of mechanisitc element is being shown for that step, i.e. "make a bond", etc.

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Rank the following alkyl halides in order of increasing SN1 reactivity. (1 – least reactive, 3 – most reactive)

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Predict the organic product of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product. Draw your answer in skeletal form. You will be graded on the product your draw from the reaction no other information is needed for this question.

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Rank the following molecules in increasing order of susceptibility to an SN1 reaction.

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Using arrows to show the flow of electrons, write a stepwise mechanism for the reaction given below. Furthermore, draw a structural representation of the latest (also known as the most product-like) transition state structure with consideration of each elementary steps in the reaction mechanism.

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Energy Diagram and Features of SN1 Reactions

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 Provide the major product for the following reaction. 

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Draw the mechanism for each of the following reactions: 

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Draw the mechanism for each of the following reactions: 

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Draw the mechanism of the following SN1 processes: 

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Draw the mechanism of the following SN1 processes: 

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Draw the mechanism of the following SN1 processes: 

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Draw the mechanism for the following SN1 reaction

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Predict the product(s) for the following reaction.

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A) Provide structures in the box to complete the Lewis acid-base reaction.  B) Create the electron arrows needed for the reactants to produce the product.

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For each of the following sets of molecules, CIRCLE the set that will undergo an S N1 reaction faster. If both react equally fast, CIRCLE BOTH. 

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How many atoms and electrons are directly involved in the bond-making and bond-breaking of the first step in an SN1 reaction?

(A) three atoms, four electrons

(B) two atoms, four electrons 

(C) three atoms, two electrons 

(D) two atoms, two electrons

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Draw the mechanism for the solvolysis of (R)-3-iodo-3-methylhexane in water, leading to a neutral organic product. Show all lone pairs, formal charges, and (where appropriate) stereochemistry.

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What set of reaction conditions would favor an SN1 reaction on 2-bromo-3-methylbutane?

a)Strong nucleophile in a protic solvent.

b) Strong nucleophile in an aprotic solvent.

c) Weak nucleophile in a protic solvent.

d) Weak nucleophile in an aprotic solvent.

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Draw the structure(s) of the product(s) that you would expect from the following reaction.

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Use A-G to label the graph with the structures given for the reaction:

 

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The hybridization state of the charged carbon in an S N1 reaction is:

  1. s
  2. sp
  3. sp2
  4. sp3
  5. sp4
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Compare the following reactions and decide which reaction in each pair would occur faster. Write your answer and concisely defend your choice. No reasoning, no points.

 

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